# What is the ordinary generating function of this series?

I need this as part of a bigger proof.

Does \begin{align} G_1 &\longleftrightarrow \Big\{\binom{n}{k}\Big\}_{2k=0}^r \notag \\ \implies G_1(z) &= (1 + z^2)^r \end{align}

Help me prove this as it would help me with the bigger problem I'm doing.

I know the ordinary generating function of $\Big\{\binom{r}{k}\Big\}_{k=0}^r$ is $(1 + z)^r$.

Here is the original problem and my working.

Find a generating function $G(z)$ for which $$[z^n]G(z) = \sum_{k_0 + 2k_1 + 4k_2 + 8k_3 + \dots =n} \binom{r}{k_0}\binom{r}{k_1}\binom{r}{k_2}\binom{r}{k_3}\dots$$

The fact that all the sum of all the indices of summation is always constant suggests that $G(z)$ is the product of several different generating functions. \begin{align} \text{The generating function of }G_1 &\longleftrightarrow \Big\{\binom{r}{k_0}\Big\}_{k_0 = 0}^r \notag \\ \implies G_1(z) &= (1+z)^r \\ \text{The generating function of }G_2 &\longleftrightarrow \Big\{\binom{r}{k_1}\Big\}_{2k_1 = 0}^r \notag \\ \implies G_2(z) &= (1+z^2)^r \\ \text{The generating function of }G_3 &\longleftrightarrow \Big\{\binom{r}{k_2}\Big\}_{4k_2 = 0}^r \notag \\ \implies G_3(z) &= (1+z^4)^r\\ \vdots \notag \\ \text{The generating function $G(z)$ can be gotten by multiplying all these functions}G(z) &= G_1(z)G_2(z)G_3(z)\dots \notag \\ &=(1+z)^r(1+z^2)^r(1+z^4)^r\dots \notag \\ \implies G(z) &=\Big(\frac{1}{1-z}\Big)^r \\ \text{Now, we have gotten the generating function. The sum is }&\binom{r+n-1}{n} \end{align}

• What is $\Big\{\binom{n}{k}\Big\}_{2k=0}^r$? – Ross Millikan Jul 3 '16 at 14:00
• @RossMillikan It's a sequence of numbers. $G_1(z)$ is the generating function of that sequence. – user230452 Jul 3 '16 at 14:02
• @RossMillikan Please help me, Ross. You can do it. – user230452 Jul 3 '16 at 14:07
• In your last, it is only $(1+z)^r$ if $r=n$. Is that true in the case you want? – Ross Millikan Jul 3 '16 at 14:17
• @RossMillikan That was a typo – user230452 Jul 3 '16 at 14:18

$$(1+z^2)^r=\sum_{k=0}^r\binom rkz^{2k}$$ hence $G_{1,2k}=\binom rk$ and $G_{1,2k+1}=0$.
• Ok, thanks. ${}$ – user230452 Jul 3 '16 at 15:26